# Vuk's question at Yahoo! Answers regarding the evaluation of an indefinite integral

#### MarkFL

Staff member
Here is the question:

How do you solve the integral of (lnx)dx/(1+x^2)^(3/2)?
I have posted a link there to this topic so the OP can see my work.

#### MarkFL

Staff member
Hello Vuk,

We are given to evaluate:

$$\displaystyle I=\int\frac{\ln(x)}{\left(x^2+1 \right)^{\frac{3}{2}}}\,dx$$

If we use integration by parts, we could let:

$$\displaystyle u=\ln(x)\,\therefore\,du=\frac{1}{x}\,dx$$

$$\displaystyle dv=\frac{1}{\left(x^2+1 \right)^{\frac{3}{2}}}\,dx$$

To find $v$, we may use a trigonometric substitution:

$$\displaystyle x=\tan(\theta)\,\therefore\,dx= \sec^2(\theta)\,d\theta$$

and we find:

$$\displaystyle v=\frac{\sec^2(\theta)}{\left(\tan^2(\theta)+1 \right)^{\frac{3}{2}}}\,d\theta$$

Using the Pythagorean identity $$\displaystyle \tan^2(\theta)+1=\sec^2(\theta)$$ we get:

$$\displaystyle v=\frac{\sec^2(\theta)}{\sec^3(\theta)}\,d\theta= \int\cos(\theta)\,d\theta=\sin(\theta)$$

Back-substituting for $\theta$, we obtain:

$$\displaystyle v=\sin\left(\tan^{-1}(x) \right)=\frac{x}{\sqrt{x^2+1}}$$

And so we have:

$$\displaystyle I=\frac{x\ln(x)}{\sqrt{x^2+1}}-\int\frac{1}{\sqrt{x^2+1}}\,dx$$

Now, using the same trigonometric substitution we used before, we have:

$$\displaystyle I=\frac{x\ln(x)}{\sqrt{x^2+1}}- \int\sec(\theta)\,d\theta$$

Let:

$$\displaystyle u=\sec(\theta)+\tan(\theta)\, \therefore\,du= \left(\sec(\theta) \tan(\theta)+\sec^2( \theta) \right)\,d\theta=$$

$$\displaystyle \sec(\theta)\left(\tan(\theta)+\sec(\theta) \right)\,d\theta=u \sec(\theta)\,d\theta\,\therefore\,\sec(\theta)\,d\theta=\frac{1}{u}\,du$$

Hence, we now have:

$$\displaystyle I=\frac{x\ln(x)}{\sqrt{x^2+1}}-\int\frac{1}{u}\,du$$

$$\displaystyle I=\frac{x\ln(x)}{\sqrt{x^2+1}}-\ln|u|+C$$

Back-substitute for $u$:

$$\displaystyle I=\frac{x\ln(x)}{\sqrt{x^2+1}}-\ln|\sec(\theta)+\tan(\theta)|+C$$

Back-substitute for $\theta$:

$$\displaystyle I=\frac{x\ln(x)}{\sqrt{x^2+1}}-\ln\left|\sqrt{x^2+1}+x \right|+C$$

And we are done.